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Solid Mechanics
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Anil Mandariya 1
CE 251 (Solid Mechanics)
Lecture-17 & 18
Torsion Circular shaft: assumptions, stress-strain
distribution, angle of twist
&
Multiple Torque and Cross-Section Areas, statically Determinate and Indeterminate
torque loaded member
Anil Mandariya 2
Lecture-17
Torsion Circular shaft: assumptions, stress-strain
distribution, angle of twist
Anil Mandariya 3
Moment of a force: The moment of a force vector results from the cross product of a position vector with a force vector.
Moment of a force system: The moment of a set S of force vectors (bound vector) FQ1, ...,FQn about a point O is defined as the sum of the moments of each force vector (bound
vector) in equation.
Moment, Torque:
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Shift theorem for the moment of a set of vectors: MS/P, the moment of a set S of force vectors about a point P, can be calculated in terms of:
MS/O, the moment of S about a point O rO/P , O’s position vector from P FS, the resultant of S
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Torque: Torque is the moment of a set S of force vectors whose resultant is zero.
A couple is a set of vectors whose resultant (sum) is 0, a torque is the moment of a couple.
A couple has a special property, namely, the moment of a couple about a point O is equal to the moment of the couple about any other point Q. As a result, a torque is not associated with a point.
All Torques are Moments, but not all Moments are Torques.
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Torsion is the twisting of an object due to an applied torque, therefore is expressed in N·m.
Examples of bars in torsion are drive shafts in automobiles, axles, propeller shafts, steering rods, and drill bits.
Torsion:
8
Assumptions: (1) Material is homogeneous and isotropic.
(2) A plane section of material perpendicular to the axis of a circular member remains plane after the torques are applied, i.e., no warpage or distortion of parallel planes normal to the axis of a member takes place.
(3) Material is linear elastic
Anil Mandariya
Torsion of circular shaft:
9
Derivation: Consider a solid cylinder with cross sectional area A
and length L. Let the cylinder be subjected to a twisting couple T
applied at the right end. An equilibrating torque acts on the left end. The vectors that represent the torque are directed
along the z axis, the centroidal axis of the shaft.
Anil Mandariya *Source: Engineering Mechanics of solids by E.P. Popove
R
11
Anil Mandariya
R
Under the action of the torque, an originally straight generator of the cylinder AB will deform into a helical curve AB*.
Because of the radial symmetry of the circular cross section and because a deformed cross section must appear to be the same from both ends of the torsion member, plane cross sections of the torsion member normal to the z axis remain plane after deformation and all radii remain straight.
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For small displacements, the torque T causes each cross section to rotate as a rigid body about the z axis (axis of the couple); this axis is called the axis of twist.
If we measure the rotation β of each section relative to the plane z = 0, the rotation β of a given section will depend on its distance from the plane z = 0.
For small deformations, assume that the amount of rotation of a given section depends linearly on its distance z from the plane z = O. Thus,
where ϴ is the angle of twist per unit length of the shaft
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Because plane sections are assumed to remain plane, the displacement component ‘w’, parallel to the z axis, is taken to be zero.
To calculate the (x, y) components ‘u’ and ‘v’, consider a cross-section distance z from the plane z = 0.
Consider a point in the circular cross section with radial distance OP and Under the deformation, radius OP rotates into the radius OP∗.
In terms of the angular displacement θ of the radius, the displacement components (u, v) are:
Expanding cos(β + ϕ) and sin(β + ϕ) and noting that x = OP cosϕ, y = OP sinϕ, we may write in the form
Restricting the deformation to be small, so sinβ ≈ β, cos β ≈ 1.
Anil Mandariya * *Source: ADVANCED MECHANICS OF MATERIALS by ARTHUR P. BORESI et. Al. 14
β
15
In terms of the angular displacement θ of the radius, the displacement components (u, v) are:
Or
On the basis of the foregoing assumptions, above Eqs. represent the displacement components of a circular shaft subjected to a twisting couple M.
As strain components are:
Anil Mandariya * *Source: ADVANCED MECHANICS OF MATERIALS by ARTHUR P. BORESI et. Al.
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So, the strain components are: ;
According to generalized Hook’s law:
From the above equations:
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They automatically satisfy compatibility equations in the absence of body forces and temperature.
They satisfy equilibrium provided the body forces are zero.
To satisfy the boundary conditions, Eqs.
must yield no forces on the lateral boundary; on the
ends they must yield stresses such that the net moment is equal to M and the resultant force vanishes.
18
R
Because the direction cosines of the unit normal to the lateral surface are (l, m, 0):
For the boundary forces As
so, or Accordingly, the boundary conditions on the lateral
boundary are satisfied.Anil Mandariya
19
Because all stress components except τyz, τxz vanish, summation of forces on the end planes yield Fx = Fy = Fz = 0.
The summation of moments with respect to the z axis yields:
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R
20
is the polar moment of inertia (J = πR4/2) of the
circular cross section.
So, Because τxy, τyz are independent of z, the stress
distribution is the same for all cross sections.
The stress vector "τ" for any point P in a cross section is given by the relation:
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21
So, torsion formula for circular shaft is:
τ is a maximum for r = R; that is, τ attains a maximum
value of ϴGR on the lateral boundary of the shaft.
Variation of shear stresses on circular shaft cross section is as following as:
τmax
Elastic hollow circular tubeElastic two co-centric solid circular tubeAnil Mandariya
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The shear stresses acting only in the plane of a cut perpendicular to the axis of the shaft.
They are acting to form a couple resisting the externally applied torques.
The shear stresses acting in the plane of the cuts taken normal to the axis of a rod cannot exist alone, but numerically, equal shear stresses must act on the axial planes (such as the planes aef and bcg in fig to fulfill the requirements of static equilibrium for an element.
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Remarks on the Torsion Formula:
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The shear stresses acting in the axial planes follow the same variation in intensity as do the shear
stresses in the planes perpendicular to the axis of the rod.
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If the shear stresses can be transformed into an equivalent system of normal stresses acting at angles of 45° with the shear stresses.
Numerically, these stresses are related to each other in the following manner: τ = σ1 = - σ2.
Therefore, if the shear strength of a material is less than its strength in tension, a shear failure takes place on a plane perpendicular to the axis of a bar.
This kind of failure occurs gradually and exhibits ductile behavior.
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If the shear stresses can be transformed into an equivalent system of normal stresses acting at angles of 45° with the shear stresses.
Numerically, these stresses are related to each other in the following manner: τ = σ1 = - σ2.
if σ1 < τ, a brittle fracture is caused by the tensile
stresses along a helix forming an angle of 45° with the bar.
Part of fracturedsandstone core specimen intorsion.
Fractured cast ironspecimen in torsion.
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Pure Torsion condition:
The total angle of twist ϕ, equal to the rate of twist times the length of the bar (that is, ϕ = ϴL), is
in which ϕ is measured in radians.
The quantity GJ/L, called the torsional stiffness (kT) of the bar or shaft, is the torque required to produce a unit angle of rotation.
The torsional flexibility (fT) is the reciprocal of the stiffness.
ϕL
ϕLkT
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Lecture-18
Multiple Torque and Cross-Section Areas, statically Determinate and Indeterminate
torque loaded member
28
D
The all internal torques and all reactions can be obtained from free-body diagrams and equations of equilibrium.
Steps for solving the problems:Step-1: Draw free-body diagrams of the given physical situation. The unknown quantities in the equilibrium equations are torques, either internal torques or reaction torques.
TAB - T =0 TAB = T
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Statically determinate torsional members:
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Step-2: Write the equations of compatibility, based upon physical conditions pertaining to the angles of twist: the compatibility equations contain angles of twist as unknowns.
ϕAB = ϕA
Step-3: Relate the angles of twist to the torques by torque-displacement relations.
ϕA
LTAB
30
The horizontal shaft AD is attached to a fixed base at D and is subjected to the torques as shown. A 44 mm diameter hole has been drilled into portion CD of the shaft. Knowing that the entire shaft is made of steel for which G = 77 GPa, determine the angle of twist at end A.
Anil Mandariya
Example-1
31
Step-1: Draw free-body diagrams of the given physical situation.
Since the shaft consists of three portions AB, BC, and CD, each of uniform cross section and each with a constant internal torque.
Portion AB
TAB – 250 = 0
or TAB = 250 N-m
BInternal torque TAB
250 N-m
A
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Solution:
32Anil Mandariya
Portion BC
TBC – 2000 - 250 = 0
or TBC = 2250 N-m
Portion CD
TCD – 2000 - 250 = 0
or TCD = 2250 N-m
B
TBC
250 N-m
A
2000 N-m
B
TCD
250 N-m
A
2000 N-m
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Step-2: Write the equations of compatibility
ϕA = ϕAB + ϕBC + ϕCD
Step-3: Relate the angles of twist to the torques by torque-displacement relations:
34CD Cross-section
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Polar moment of inertia of AB, BC & CD cross-section are:
AB Cross-section
15 mm
BC Cross-section
30 mm
30 mm
22 mm
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Step-3: Calculate the angles of twist to the torques by torque-displacement relations:
36
Two solid steel shafts are connected by the gears shown. Knowing that for each G = 77.22 x 109 Pa and that the allowable shearing stress is 55.16 MPa, determine (a) the largest torque To that may be applied to and A of the shaft AB, (b) the corresponding angle through which end A of shaft AB rotates.
Anil Mandariya
Assignment:
37
A solid steel shaft ABC of 50 mm diameter is driven at A by a motor that transmits 50 kW to the shaft at 10 Hz. The gears at B and C drive machinery requiring power equal to 35 kW and 15 kW, respectively.
Compute the maximum shear stress τmax in the shaft
and the angle of twist ϕAC between the motor at A and the gear at C. (Use G = 80 GPa).
Anil Mandariya
Example-2
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Step-1: Determining the torques applied to the shaft at A by the motor.
Since the motor supplies 50 kW at 10 Hz, it creates a torque TA at end A of the shaft
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Step-2: Determining the torques applied to the different segments of shaft.
Segment AB
TAB – TA = 0
TAB – 796 = 0 TAB = 796 kN-m
By
y’TAB
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Step-3: Determine the shear stresses and angles of twist of segments.
Segment AB
By
y’TAB
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Segment BC:
Thus the maximum shear stress in shaft is in segment AB.
C
TBC = 239 kN-my’
y239 kN-m B
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The total angle of twist between the motor at A and the gear at C is:
As both parts of the shaft twist in the same direction, and therefore the angles of twist are added.
Steps for solving the problems:Step-1: Draw free-body diagrams of the given physical
situation. The unknown quantities in the equilibrium equations are torques, either internal torques or reaction torques.
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Statically indeterminate torsional members:
46Anil Mandariya
ϕ1
ϕo
Step-3: Equation of compatibility Separate the bar from its support at any one end (say B)
and obtain a bar that is fixed at other end (say A) and free at one end (say B). When the load To or all external torques acts alone, they produces an angle of twist at end B that we denote as ϕo. Similarly, when the only reactive torque TB acts alone, it produces an angle ϕ1.
The angle of twist at end B in the original bar, equal to the sum of ϕo and ϕ1, is zero. Therefore, the equation of compatibility is:
ϕo + ϕ1 = 0
47Anil Mandariya
Step-3: Torque-displacement equations The angles of twist ϕo and ϕ1 can be expressed in terms
of the torques To and TB by referring. The equations are as follows:
Step-4: Substitute the angles of twist into the compatibility equation.
48Anil Mandariya
Step-5: Solution of equations obtained from step-4 and equilibrium equation and determine the value of internal torque reactions.
Step-6: Maximum shear stresses:
Step-7: Angle of rotation: The angle of rotation fC at section C is equal to the
angle of twist of either segment of the bar, since both segments rotate through the
same angle at section C.
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